# Rolle's Theorem Questions and Examples

Rolle's theorem is a special case of the mean value theorem. It is discussed here through examples and questions.

## Rolle's Theorem

Rolle's theorem is the result of the mean value theorem where under the conditions:f(x) be a continuous functions on the interval [a, b]

and differentiable on the open interval (a, b) , there exists at least one value c of x such that f '(c) = [ f(b) - f(a) ] /(b - a).

Now if the condition f(a) = f(b) is satisfied, then the above simplifies to : f '(c) = 0.

In other words, under the above three conditions we can always find a tangent to the curve of f that is horizontal (slope = f'(c) = 0).

Rolle's Theorem

If f(x) is

1) a continuous function on the interval [a, b]

2) differentiable on the open interval (a, b)

3) and f(a) = f(b),

then there is at least one value c of x in the interval (a, b) such that

f '(c) = 0

1) a continuous function on the interval [a, b]

2) differentiable on the open interval (a, b)

3) and f(a) = f(b),

then there is at least one value c of x in the interval (a, b) such that

f '(c) = 0

## Example 1

The graph of f(x) = - x^{2}+ 6x - 6 for 1 ≤ x ≤ 5 is shown below. f(1) = f(5) = - 1 and f is continuous on [1 , 5] and differentiable on (1 , 5) hence, according to Rolle's theorem, there exists at least one value of x = c such that f '(c) = 0.

f '(x) = - 2 x + 6

f '(c) = - 2 c + 6 = 0

Solve the above equation to obtain

c = 3

Therefore at x = 3 there is a tangent to the graph of f that has a slope equal to zero (horizontal line) as shown in figure 1 below.

## Example 2

The graph of f(x) = sin(x) + 2 for 0 ≤ x ≤ 2π is shown below. f(0) = f(2π) = 2 and f is continuous on [0 , 2π] and differentiable on (0 , 2π) hence, according to Rolle's theorem, there exists at least one value ( there may be more than one! ) of x = c such that f '(c) = 0.f '(x) = cos(x)

f '(c) = cos(c) = 0

The above equation has two solutions on the interval [0 , 2π]

c

_{1}= π/2 and c

_{2}= 3π/2.

Therefore both at x = π/2 and x = 3 π/2 there are tangents to the graph that have a slope equal to zero (horizontal line) as shown in figure 2 below.

## Example 3

Function f in figure 3 does not satisfy Rolle's theorem: although it is continuous and f(-1) = f(3), the function is not differentiable at x = 1 and therefore f '(c) = 0 with c in the interval (-1 , 3) is not guaranteed. In fact it is easy to see that there is no horizontal tangent to the graph of f on the interval (-1 , 3).## Question 1

Which of the functions given below satisfy all three conditions of Rolle's theorem?a) f(x) = cos(x) , for x in [0 , 2π]

b) g(x) = |x - 2| , for x in [0 , 4]

c) h(x) = 1 / x

^{2}, for x in [-1 , 1]

d) k(x) = |sin(x)| , for x in [0 , 2π]

### solution to question 1

a)f(0) = 1 and f(2π) = 1 therefore f(0) = f(2π)

f is continuous on [0 , 2π]

Function f is differentiable in (0 , 2π)

Function f satisfies all conditions of Rolle's theorem

b)

function g has a V-shaped graph with vertex at x = 2 and is therefore not differentiable at x = 2.

Function g does not satisfy all conditions of Rolle's theorem

c)

Function h is undefined at x = 0.

Function h does not satisfy all conditions of Rolle's theorem.

d)

k(x) = |sin(x)| , for x in [0 , 2π]

The graph of function k is shown below and it shows that function k is not differentiable at x = π.

Function k does not satisfy all conditions of Rolle's theorem.

## Question 2

Check that function f(x) = x^{ 2}- 4 x + 3 on the interval [1 , 3] satisfies all conditions of Rolle's theorem and then find all values of x = c such that f '(c) = 0.

### solution to question 2

f is a polynomial function and is therefore continuous on the interval [1 , 3] and differentiable on the interval (1 , 3). Also f(1) = f(3) = 0 and therefore function f satisfies all three conditions of Rolle's theorem and there is at least one value of x = c such that f '(c) = 0.f '(x) = 2 x - 4

f '(c) = 2 c - 4 = 0

Solve for c to obtain

c = 2.

The graph below shows f , f ' and the tangent at x = c = 2 is horizontal; and f '(2) = 0

## Question 3

Check that function g(x) = cos(x) on the interval [- π/2 , 3π/2] satisfies all conditions of Rolle's theorem and then find all values x = c such that g '(c) = 0.### solution to question 3

Function g is a cosine function and is therefore continuous on the interval [- π/2 , 3π/2] and differentiable on the interval(-π/2 , 3π/2). Also g(- π/2) = g(3π/2) = 0 and therefore function g satisfies all three conditions of Rolle's theorem and there is at least one value of x = c such that f '(c) = 0.

g '(x) = - sin(x)

g '(c) = - sin(c) = 0

Solve for c to obtain

c = nπ , n = 0,~+mn~ 1 , ~+mn~ 2 , ...

Solutions on the interval [- π/2 , 3π/2] are

c

_{1}= 0 and c

_{2}= π

The graph below shows g , g ' and the tangents at x = c

_{1}= 0 and x = c

_{2}= π are horizontal; and g '(0) = 0 and g '(π) = 0

__More References and links__

problems related to the mean value theorem Continuous Functions in Calculus

Non Differentiable Functions

calculus